What is (\partial)/(\partial y)(x^2y+zcos(y)) ?

The answer to (\partial)/(\partial y)(x^2y+zcos(y)) is x^2-zsin(y)

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(\partial)/(\partial y)(x^2y+zcos(y))

\frac{\partial}{\partial y} (x^{2} y + z cos (y))

x^{2} - z sin (y)

Solution steps

\frac{\partial}{\partial y} (x^{2} y + z cos (y))

Treat

x, z

as constants

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial y} (x^{2} y) + \frac{\partial}{\partial y} (z cos (y))

\frac{\partial}{\partial y} (x^{2} y) = x^{2}

\frac{\partial}{\partial y} (z cos (y)) = - z sin (y)

= x^{2} - z sin (y)

\frac{d}{d x} (sin (- 2 ln (1 - 2 x)))

\frac{x}{x ^{2} - 3}

y^{^{'}} + \frac{1}{x} y = 4 (x + 7)^{2}

4 x (x^{2} - 1)

\frac{d y}{d t} = 16 - \frac{y}{250 + 0 t}

What is (\partial)/(\partial y)(x^2y+zcos(y)) ?
The answer to (\partial)/(\partial y)(x^2y+zcos(y)) is x^2-zsin(y)